Question: Simplify; express your answer in exponential form. Assume $y\neq 0, r\neq 0$. $\dfrac{{(y^{-4}r^{-2})^{-3}}}{{(y^{5}r^{-4})^{-2}}}$
Answer: To start, try simplifying the numerator and the denominator independently. In the numerator, we can use the distributive property of exponents. ${(y^{-4}r^{-2})^{-3} = (y^{-4})^{-3}(r^{-2})^{-3}}$ On the left, we have ${y^{-4}}$ to the exponent ${-3}$ . Now ${-4 \times -3 = 12}$ , so ${(y^{-4})^{-3} = y^{12}}$ Apply the ideas above to simplify the equation. $\dfrac{{(y^{-4}r^{-2})^{-3}}}{{(y^{5}r^{-4})^{-2}}} = \dfrac{{y^{12}r^{6}}}{{y^{-10}r^{8}}}$ Break up the equation by variable and simplify. $\dfrac{{y^{12}r^{6}}}{{y^{-10}r^{8}}} = \dfrac{{y^{12}}}{{y^{-10}}} \cdot \dfrac{{r^{6}}}{{r^{8}}} = y^{{12} - {(-10)}} \cdot r^{{6} - {8}} = y^{22}r^{-2}$